I was given the following exercise: let $S$ be $x^2 +y^2-z^2=1$.

  1. show that for every real number $t$ the line $l_t$ $$(x-z)\cos t=(1-y)\sin t,\quad (x+z)\sin t=(1+y)\cos t$$ is contained in $S$;
  2. show that every point of $S$ is contained in one and only one of the above lines;
  3. use this remark to parametrize $S$.

My approach was to observe that $l_t = l_{t+k\pi}$; then I defined $t:=\arctan\left(\frac{1+y}{x+z}\right)$ and showed that $$p=(x,y,z) \in S \iff p \in l_t\text{,}$$ obviously assuming $t\neq \pm \frac{\pi}{2}, y\neq 1, x\neq -z$. The 'only one' part was not a problem. Finally I wrote $l_t$ in parametric form obtaining a parametrization for $S$.

My question is: is there a way to solve the exercise so that the parametrization obtained for $S$ is unique? Unfortunately in my solution I have to consider an atlas of parametrizations, since only one parametrization covers $S$ minus a line (e.g. $y=1, x=-z$).

Thank you.

  • $\begingroup$ "contained in one and only one of the above lines": this is a little contradictory as $t$ is unbounded and the lines are repeated infinitely many times. $\endgroup$
    – user65203
    Jun 12, 2020 at 9:25

2 Answers 2


Your lines $l_t$ are presented in a strange way. Solving the two equations for $x$ and $y$ gives $$l_t:\qquad\eqalign{x&=\sin(2t)+\cos(2t)\> z,\cr y&=-\cos(2t)+\sin(2t)\> z\ .\cr}$$ I'm taking the freedom to replace your parameter $t$ that "numbers" the lines $l_t$ by $$\phi:=2t-{\pi\over2}\quad(0\leq t<\pi),\qquad{\rm resp.}\qquad t={1\over2}\left(\phi+{\pi\over2}\right)\quad(0\leq\phi<2\pi)\ .$$ I'm then talking about the lines $$\ell_\phi: \quad z\mapsto\bigl(\cos\phi-\sin\phi \>z,\ \sin\phi+\cos\phi \>z,\ z\bigr)\qquad(-\infty<z<\infty)\tag{1}$$ with $0\leq\phi<2\pi$. When $z=0$ the line $\ell_\phi$ passes through the point $(\cos\phi,\sin\phi),0)$ of the unit circle in the $(x,y)$-plane. In fact the projection $$\ell'_\phi: \quad z\mapsto\bigl(\cos\phi-\sin\phi \>z,\ \sin\phi+\cos\phi \>z,\ 0\bigr)\qquad(-\infty<z<\infty)$$ of the $\ell_\phi$ into the $(x,y)$-plane is just the tangent to the unit circle, $\phi$ being the polar angle of the point of tangency. When I change the value of $\phi$ to a $\phi'\ne\phi$ (modulo $2\pi$) I obtain a different tangent, and it is then geometrically obvious that the lines $\ell_\phi$ and $\ell_{\phi'}$ don't intersect in space.

The new parametrization of $S$ now comes for free from $(1)$. We just write $$S:\quad (\phi,z)\mapsto\left\{\eqalign{x&=\cos\phi-\sin\phi \>z\cr y&=\sin\phi+\cos\phi \>z\cr z&=z\cr}\right.\qquad\qquad(\phi\in{\mathbb R}/(2\pi), \ -\infty<z<\infty)\ .$$

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I guess that the expected solution is by parameterizing the points along the lines.

Taking $y$ as a free variable,

$$x-z=(1-y)\tan(t)\\x+z=(1+y)\cot(t)$$ give you a parameterization in terms of $y$ and $t$. If $t$ is contrained to $[0,\pi)$, the parameterization is unique from the result 2. and the fact that we intersect a straight line with a plane.

(The special cases such that $\tan(t)$ or $\cot(t)$ are undefined correspond to the two lines $y=\pm1, x=\pm z$.)


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